# Solve system of linear equations, using matrix method: $2x-y=-2$ $3x+4y = 3$

Toolbox:
• A matrix is said to be singular if |A|= 0.
• A matrix is said to be invertible if |A|$\neq 0$.
• If A is a non-singular matrix,AX=B,then $X=A^{-1}B.$
• Using this we can solve the system of equation which has unique solution.
This can be written in the form AX=B.

$\begin{bmatrix}2 & -1\\3 & 4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-2\\3\end{bmatrix}$

Where $A=\begin{bmatrix}2 & -1\\3 & 4\end{bmatrix}\;X=\begin{bmatrix}x\\y\end{bmatrix}\;B=\begin{bmatrix}-2\\3\end{bmatrix}$

Let us evaluate $|A|=2\times 4-(-1)\times 3$

$\qquad\qquad\qquad=8+3=11.$

$|A|\neq 0.$

Hence it is non singular.

Therefore inverse exists.

Now let us find out adj A.

This can be obtained by interchanging the element of $a_{11}$ and $a_{22}$ and interchanging th symbols of elements of $a_{21}$ and $a_{12}$

Therefore $adj A=\begin{bmatrix}4 & 1\\-3 & 2\end{bmatrix}$ and |A|=11.

$A^{-1}=\frac{1}{11}\begin{bmatrix}4 & 1\\-3 & 2\end{bmatrix}$

Matrix multiplication can be obtained by multiplying the rows of A with the column of B.

Now AX=B$\Rightarrow X=A^{-1}B.$

$x=\frac{1}{11}\begin{bmatrix}4 & 1\\-3 & 2\end{bmatrix}\begin{bmatrix}-2\\3\end{bmatrix}$

$\begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{11}\begin{bmatrix}-8+3\\6+6\end{bmatrix}=\frac{1}{11}\begin{bmatrix}-5\\12\end{bmatrix}$

$\Rightarrow \begin{bmatrix}x\\y\end{bmatrix}= \begin{bmatrix}-5/11\\12/11\end{bmatrix}$

Hence x=-5/11 and y=12/11.

edited Feb 27, 2013